Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $k = \dfrac{a - 4}{a + 1} \times \dfrac{-10a^2 - 20a}{a^2 - 2a - 8} $
Answer: First factor the quadratic. $k = \dfrac{a - 4}{a + 1} \times \dfrac{-10a^2 - 20a}{(a - 4)(a + 2)} $ Then factor out any other terms. $k = \dfrac{a - 4}{a + 1} \times \dfrac{-10a(a + 2)}{(a - 4)(a + 2)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (a - 4) \times -10a(a + 2) } { (a + 1) \times (a - 4)(a + 2) } $ $k = \dfrac{ -10a(a - 4)(a + 2)}{ (a + 1)(a - 4)(a + 2)} $ Notice that $(a + 2)$ and $(a - 4)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -10a\cancel{(a - 4)}(a + 2)}{ (a + 1)\cancel{(a - 4)}(a + 2)} $ We are dividing by $a - 4$ , so $a - 4 \neq 0$ Therefore, $a \neq 4$ $k = \dfrac{ -10a\cancel{(a - 4)}\cancel{(a + 2)}}{ (a + 1)\cancel{(a - 4)}\cancel{(a + 2)}} $ We are dividing by $a + 2$ , so $a + 2 \neq 0$ Therefore, $a \neq -2$ $k = \dfrac{-10a}{a + 1} ; \space a \neq 4 ; \space a \neq -2 $